The value of `lim_(x to oo)((1+3x)/(2+3x))^((1-sqrt(x))/(1-x))` is

Options

A. 0
B. `-1`
C. `e`
D. `1`

Answer

To solve the limit \( \lim_{x \to \infty} \left( \frac{1 + 3x}{2 + 3x} \right)^{\frac{1 - \sqrt{x}}{1 - x}} \), we will break it down into manageable parts.

### Step 1: Evaluate the limit of the base
First, we will evaluate the limit of the base \( \frac{1 + 3x}{2 + 3x} \) as \( x \to \infty \).

\[ \lim_{x \to \infty} \frac{1 + 3x}{2 + 3x} \]

This is an indeterminate form \( \frac{\infty}{\infty} \). To simplify, we can divide the numerator and the denominator by \( x \):

\[ = \lim_{x \to \infty} \frac{\frac{1}{x} + 3}{\frac{2}{x} + 3} \]

As \( x \to \infty \), \( \frac{1}{x} \to 0 \) and \( \frac{2}{x} \to 0 \):

\[ = \frac{0 + 3}{0 + 3} = 1 \]

### Step 2: Evaluate the limit of the exponent
Next, we will evaluate the limit of the exponent \( \frac{1 - \sqrt{x}}{1 - x} \) as \( x \to \infty \).

\[ \lim_{x \to \infty} \frac{1 - \sqrt{x}}{1 - x} \]

This is also an indeterminate form \( \frac{-\infty}{-\infty} \). We can simplify it by multiplying the numerator and denominator by \( -1 \):

\[ = \lim_{x \to \infty} \frac{\sqrt{x} - 1}{x - 1} \]

Now, we can divide the numerator and denominator by \( x \):

\[ = \lim_{x \to \infty} \frac{\frac{\sqrt{x}}{x} - \frac{1}{x}}{1 - \frac{1}{x}} = \lim_{x \to \infty} \frac{\frac{1}{\sqrt{x}} - 0}{1 - 0} \]

As \( x \to \infty \), \( \frac{1}{\sqrt{x}} \to 0 \):

\[ = 0 \]

### Step 3: Combine the results
Now we combine the results from Steps 1 and 2:

\[ \lim_{x \to \infty} \left( \frac{1 + 3x}{2 + 3x} \right)^{\frac{1 - \sqrt{x}}{1 - x}} = 1^0 \]

Since any number raised to the power of 0 is 1, we conclude:

\[ \lim_{x \to \infty} \left( \frac{1 + 3x}{2 + 3x} \right)^{\frac{1 - \sqrt{x}}{1 - x}} = 1 \]

### Final Answer
Thus, the value of the limit is:

\[ \boxed{1} \]